Chi-square test for genetics

Work out the expected count for each class from the ratio and the total, then add up (O − E)² / E across the classes. Mendel counted 705 purple and 224 white pea flowers, a total of 929; a 3 : 1 ratio predicts 696.75 and 232.25, so χ² = 0.098 + 0.293 = 0.391. That is below the critical value of 3.841 for 1 degree of freedom, so the data fit 3 : 1 and the hypothesis is not rejected.

The critical value is the largest χ² that chance alone would produce 95% of the time when the hypothesis is true (α = 0.05). It depends on the degrees of freedom: 3.841 for df = 1, 5.991 for df = 2, and 7.815 for df = 3. If your χ² is at or below it, the deviation is what sampling could produce, so you do not reject the ratio; the tool also reports the exact p-value, 0.532 for Mendel’s flower data.

Degrees of freedom equal the number of phenotype classes minus one, because once the total is fixed the last class count is not free to vary. A 3 : 1 or 1 : 1 test has df = 1, a 1 : 2 : 1 test has df = 2, and a 9 : 3 : 3 : 1 dihybrid test has df = 3. Mendel’s dihybrid counts of 315 : 108 : 101 : 32 give χ² = 0.47 against a critical value of 7.815 (p ≈ 0.93), an excellent fit; for reliable results every expected count should be at least 5.